Відео

Too many steps, just apply the definition of logarithm: it is the exponent to be given to the base to obtain the argument, so x= Log 200, x=3.010...

X = log 333/log 9 = 2.643 (approx)

Х=3

80(9^x-1)=20 9^x-1=1/4 3^(2x-2)=2^(-2) 2x-2=-2log_3(2) x=1-log_3(2) x~= 1- log2/log3 = 1-0.301/0.477 =0.176/0.477 about 0.4

Wtf? It has to be 81, doesn't it?

1+ log 37/ log 9

a=3 b=0 это на вскидку посмотрим, что получится в итоге.😅

I woud add the first and the third equation and subtract the second equation.. Tht gives 2=2 i.e. a=1. With this informtion,we cn get from the first equation b=0 and from the third equation c=2. I think, my soutionis easier than the one in the video ...

a=1 b=0 c=2 Ez 10 seconds mai answer soch liya maine

9^(n+1)-9^(n-1)=20 9•9ⁿ-[(9ⁿ)/9]=20 81•9ⁿ-9ⁿ=20 80•9ⁿ=20 9ⁿ=¼ 9ⁿ=4^-1 3²ⁿ=4^-1 2n=log_3(4^-1) 2n=log_3(2^-2) 2n=-2[log_3(2)] n=-log_3(2) ❤

Adding (1)+(2)+(3) is unnecessary. Why make it so complicated? (1)-(2)=> a-c=-1 (5) (3)+(5)=> 2a = 2 => a= 1 subs a=1 into (1) => b=0 subs a=1 into (3) => b=2

A= 1 B =0 C = 2

A-1, B-0, C-2

√15√√15√√√15=225^k 15^1/2 .15^1/4. 15^1/8=(15^2)^k 15^1/2+1/4+1/8=15^2k 15^7/8 = 15^2k Since bases are equal, powers are also equal. 2k=7/8 K = 7/16

√5√√5√√√5=125^k 5^1/2 .5^1/4. 5^1/8 =125^k 5^7/8 = (5^3)^k 5^7/8=5^3k 3k = 7/8 K= 7/24

Your 'x' is 45, '2' easy!

Deviam habituar-se a desenhar SEMPRE primeiro o traço de fração e depois o numerador e o denominador para assim ficar tudo alinhado

Messed up twice.. learn before you try to teach. -multiplication and division are equal. You go left to right, so you start with 4/2 -even if multiplication was to go first, 2(1/2)≠5/2 (which I genuinely have no f***ing clue on how you got to), 2(1/2)=2/2=1 Solution: 4/2(1/2)=? 2(1/2)=? 2/2=? ?=1

2

35 is not 7.5

bird noise is irritating

glad you explained it.

The multiplication dot was so low that I confused 2•5 with 2.5

This is in no way a "Nice ... Simplification"! It is simply arithmetic! The real 'problem' here is that the problem posed does not have a "Nice" solution. It is usual that the solutions to this kind of problem are in the form of, say, x^n eg (32^2)-(8^3) which yields the result 2^9. 🍌🤨

Its wrong. The real answer is 62220 2×3=6 not 9

Is this supposed to be a good solution?

Desde cuándo 3×2 son 9

Excellent 😢 9^k ,= 45 k = log base9 45 =(log 45)/log9 Use calculator to find answer 4 lines Do you really know maths. Read the comments and learn

Just take log on both sides and you're done ⁹

Unfathomable and rather pointless.

To be fair though, it should also be noted that when you distributed the -4 early on in the problem, you ended up with -4x-12....... And should have had -4x+12. Had you done that, then your determinant most likely would have had imaginary solutions.

Two wrongs makes it right

Rethink this problem a bit. x^2-4x-12= (x+2)(x-6).... These are real root situation. x=-2 and x=6....... They are NOT imaginary. Where your quadric equation went wrong was in the determinant. INSIDE THE DETERMINANT: Think of it this way ----------> 16 take away (4*1*(-12))...... or 16-(-48)=64. Your actually taking the negative away when done correctly. That makes it positive. The determinant is the sq of 64, which equals 8. (4, plus or minus 8)/2...... Which simplifies to 2, plus or minus 4. Result x=6 and x=-2

Good

Bằng 4

No se lo que g, gt y la tercera llave no se ve...

Just stuck 4 in at start and it fits….. I must be a genius

😂

i dont get it

A simpler way to solve is to take ln on both sides resulting in x(ln 64) = 1/2(ln 16 + 1/2ln8). ln2 cancels out on both sides

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